231. Power of Two

easyAsked at Intel

Given an integer n, return true if it is a power of two. Intel interviewers ask because the one-line bit trick (n > 0 && (n & (n-1)) === 0) tests whether you've internalized the binary structure of powers of two — a property that shows up everywhere in cache-line alignment, ring buffers, and address arithmetic.

By Alex Chen, Founder, InterviewChamp.AI · Last verified 2026-05-19

Source citations

Public interview reports confirming this problem appears in Intel loops.

Glassdoor (2026-Q1)— Intel SWE-new-grad phone-screen reports list 'is power of two' as a 5-minute warm-up.
GeeksforGeeks (2025-07)— Intel Interview Experience archives include this in 'must-know bit tricks' for hardware-SWE rounds.
LeetCode discuss (2025-10)— Intel-tagged company-frequency list pairs this with reverse-bits and Hamming weight.

Problem

Given an integer n, return true if it is a power of two. Otherwise, return false. An integer n is a power of two, if there exists an integer x such that n == 2^x.

Constraints

-2^31 <= n <= 2^31 - 1

Examples

Example 1

Input

n = 1

Output

true

Explanation: 2^0 = 1.

Example 2

Input

n = 16

Output

true

Explanation: 2^4 = 16.

Example 3

Input

n = 3

Output

false

Example 4

Input

n = -16

Output

false

Explanation: Negative numbers are never powers of two by the problem's definition (x must produce a positive 2^x).

Approaches

1. Repeated division (brute force)

While n is even and greater than 1, divide by 2. If you end at exactly 1, it was a power of two.

Time: O(log n)
Space: O(1)

function isPowerOfTwoBrute(n) {
  if (n <= 0) return false;
  while (n > 1) {
    if (n % 2 !== 0) return false;
    n = Math.floor(n / 2);
  }
  return true;
}

Tradeoff: Works on any base by swapping the divisor. Logarithmic time. Easy to read but not the answer Intel wants.

2. Bit trick (optimal)

A positive power of two has exactly one set bit. n & (n - 1) clears the lowest set bit; if that result is 0, n had exactly one set bit.

Time: O(1)
Space: O(1)

function isPowerOfTwo(n) {
  return n > 0 && (n & (n - 1)) === 0;
}

Tradeoff: Single expression, branch-free after the n > 0 guard, no loop. This is the canonical answer — anything else on a power-of-two question signals lack of bit-arithmetic fluency.

Intel-specific tips

Intel cares about this because power-of-two checks are everywhere in systems code — buffer sizes for ring buffers (so % becomes & (mask)), DMA alignment, cache-line indexing, FFT sizes. Mention 'this is the same trick that lets ring buffers use bitwise-AND instead of modulo' or 'this is why ALIGN-UP-to-power-of-two is one op in firmware'. That context elevates the answer from 'I know a trick' to 'I know why the trick matters'.

Common mistakes

Forgetting the n > 0 guard. For n = 0, the expression 0 & (-1) is 0, so the trick incorrectly returns true.
Including negative powers of two by accident — the problem definition restricts x to non-negative; -16 is not a power of two even though |-16| is.
Using `Math.log2(n) % 1 === 0` — floating-point comparison; fails for large n due to precision (2^30 returns weird results in some engines).

Follow-up questions

An interviewer at Intel may pivot to one of these next:

Power of Three (LC 326) — no bit trick; use the largest power of 3 in int range and check divisibility.
Power of Four (LC 342) — power of two AND set bit is at an even position: `(n & 0xaaaaaaaa) === 0`.
Given n, return the next power of two >= n. (Hint: round up via OR-shift chain or use clz.)

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Output

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FAQ

Why does (n & (n - 1)) === 0 test for exactly one set bit?

If n has exactly one set bit at position k, then n - 1 has zeros at position k and ones at all lower positions. ANDing those two together produces zero. If n has more than one set bit, subtracting 1 only clears the lowest set bit and below; ANDing keeps the higher set bits intact, so the result is nonzero.

Why does Intel ask such a 'simple' question?

It's a 30-second filter: if you fumble it, you don't get to the harder rounds. The Intel hardware-SWE rubric explicitly wants candidates who reach for bit tricks reflexively because that's the day-to-day vocabulary in firmware and driver work.

Free learning resources

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